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 Calculus in python
#1
I don't have a great understanding of calculus (or python) but I need to find the average value of a curve for a project.  I've been using a widget from wolframalpha.com.  Can I duplicate that in python?

Here's the widget:

http://www.wolframalpha.com/widgets/view...dcbe645198
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#2
The best I could find with a quick Google for the symbolic solving of integrals was sympy (http://www.scipy-lectures.org/advanced/sympy.html). You could of course do a simulation of the limit (dividing the area under the curve into a bunch of small rectangles, and calculating the total area of the rectangles), but that may not suit your needs.
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#3
sympy has support for both indefinite and definite integrals:
Output:
In [1]: from sympy import integrate, Symbol, init_printing In [2]: init_printing(use_unicode=True) In [3]: x = Symbol('x') In [4]: integrate(x**2, x) Out[4]:  3 x ── 3 In [5]: integrate(x**2, (x, 1, 2)) Out[5]: 7/3
Or you can use numerical integration provided by SciPy.
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#4
I don't see how the range ("from" and "to" on the wolframalpha.com widget) should be specified.  Here's my formula:

f(x) = ( a * b ) / ( c * 1.05^x )
from = 1
to = b

http://www.wolframalpha.com/widgets/view...dcbe645198

I will have values for a, b, and c.
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#5
You can simplify your function as k * 1.05^(-x), where k = a * b / c. ( mathjax support would be handy here)
Lets suppose that k = 5 and you want to integrate over interval [1,5].

You can use scipy's quad() - its arguments are function's definition (as a python function) and lower and upper limit:
Output:
In [1]: from scipy import integrate In [2]: k, a, b = 5, 1, 5 In [3]: integrate.quad(lambda x: k * 1.05**(-x), a, b) Out[3]: (17.30418300357802, 1.9211502392899616e-13)
output is the result and an estimation of error.

You can use sympy's integrate(), where first argument is a function definition and second is a tuple, with elements being variable you integrate by, lower limit, upper limit. With sympy you can write your function almost "naturally", but you need to specify that x has a special meaning (a symbol). If you are interested only in numerical result, scipy's quad is probably easier to use.
Output:
In [4]: from sympy import integrate, Symbol In [5]: x = Symbol('x') In [6]: integrate(k * 1.05**(-x), (x, a, b)) Out[6]: 17.3041830035780
And finally, you can use a piece of paper (and a brain) and realize that the primitive function of k * 1.05^(-x) is -k / log(1.05) * 1.05^(-x). After that its just
Output:
In [7]: from math import log In [8]: -k / log(1.05) * ( 1.05**-b - 1.05**-a ) Out[8]: 17.304183003578018
As python is not a domain-specific language for math/symbol manipulation, using python for calculus would require atleast basic understanding of python and basic understanding of a calculus, you likely need to spend some effort and study both a little.
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#6
I'm getting a different output from the wolframalpha.com widget compared to your functions above.  With:

k, a, b = 5, 1, 5

and:

k = a * b / c

I get 4.32605 from the widget by using:

f(x) = ( a * b ) / ( c * 1.05^x )
from = 1
to = b

Could you point out my mistake and/or misunderstanding?
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#7
Hmm,  I was little unclear with notation. After first line i just stopped to care about original values of a, b, c - they were used only to compute k and forgotten (with k = a * b / c  function k * 1.05^(-x) is same as  (a*b)/c * 1.05^(-x) )

And rest was just integration of function k * 1.05^(-x) on interval [a, b], whatever values of k, a, b are. And for integral of 5 * 1.05^(-x) on interval [1, 5] wolfram alpha gives same result (17.3042) - just fill there 5 * 1.05^(-x), from 1, to 5.
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#8
I have to confess I'm a bit lost.  What would you say is the simplest python expression of what I've been doing in the Wolfram Alpha widget?
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#9
I think that something like
integrate.quad(lambda x: k * 1.05**(-x), a, b)
is probably the simplest way possible - to compute a definite integral you need to have a function to integrate, a lower bound, an upper bound. And you must express that you want to compute the definite integral from these three things. So I am not sure if it even can be significantly simpler ...

Anyway, I can try to rewrite what I think you tried to compute - just "verbatim" rewrite without any changes
from scipy import integrate

a, b, c = 2, 5, 2     # constants 

def my_func(x):                      # "defining" your function
    f_x = a * b / ( c * 1.05**x )    # value of f(x)
    return f_x

print( integrate.quad(my_func, 1, b) )  # you want to integrate function my_func on interval [1, b] and print result
And output of it is:
Output:
(17.30418300357802, 1.9211502392899616e-13)
I hope that it helps. And if not, perhaps someone else would volunteer ...
Larz60+ likes this post
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#10
Well, my goal is to get the same output from a python expression that I get from Wolfram Alpha but I'm getting different output.  With:

a, b, c = 2, 5, 2

I get ~17.30 from this (a one-liner is better for my application):

integrate.quad(lambda x: (a*b/c) * 1.05**(-x), 1, b)

I get ~4.32 from Wolfram Alpha by using these inputs on the page (with a, b, and c replaced with their values above):

f(x) = ( a * b ) / ( c * 1.05^x )
from = 1
to = b

http://www.wolframalpha.com/widgets/view...dcbe645198

Your function seems to be giving me exactly 4x what Wolfram Alpha is giving me.  What am I missing?

BTW, I think your code should read the following with 1 as the lower bound instead of a:

integrate.quad(lambda x: k * 1.05**(-x), 1, b)
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